Problem: Find $\lim_{x\to 0}\dfrac{\sin(x)}{5x+3\sin(x)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $\dfrac18$ (Choice C) C $\dfrac13$ (Choice D) D The limit doesn't exist.
Explanation: Substituting $x=0$ into $\dfrac{\sin(x)}{5x+\sin(x)}$ results in the indeterminate form $\dfrac{0}{0}$. Furthermore, as the expression involves mixed function types, it's not possible to manipulate it algebraically in a way that will help us find the limits. Therefore, we should use L'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to 0}\dfrac{\sin(x)}{5x+3\sin(x)} \\\\ &=\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[\sin(x)]}{\dfrac{d}{dx}[5x+3\sin(x)]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to 0}\dfrac{\cos(x)}{5+3\cos(x)} \\\\ &=\dfrac{\cos(0)}{5+3\cos(0)} \gray{\text{Substitution}} \\\\ &=\dfrac{1}{8} \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[\sin(x)]}{\dfrac{d}{dx}[5x+3\sin(x)]}$ actually exists. In conclusion, $\lim_{x\to 0}\dfrac{\sin(x)}{5x+3\sin(x)}=\dfrac18$.